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I was reading a comment on MathOverflow and did not understand what was being suggested. I also don't understand Anton's response:

I initially thought Regenbogen's comment was being applied to the derivative of the dirac distribution, i.e. he wanted to extend this continuous linear functional from $D$ to $C$. But I don't understand what dominating semi-norm (sublinear functional) he had in mind... And I don't understand Anton's response - How are the derivatives of the delta distribution not bounded linear functionals on $D$? They seem to have operator norm equal to 1.

Then I thought perhaps he literally was referring to the measure mentioned earlier, i.e. a Radon measure acting as a distribution on $C$ and trivially extending to $D$... but this also seems to be a bounded linear functional on both $C$ and $D$.

Can someone explain what was being suggested, and how Anton's response makes sense?

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Check your $C$ and $D$. Anton said that $\delta'$ is a linear functional on $D$, but cannot be extended to a bounded linear functional on $C$. This is natural, as $\delta'(f) = f'(0)$ and you can find continuous functions with compact support that is not differentiable at the origin. – Willie Wong Dec 23 '10 at 20:33
Now, about Rogenbogen's comment: you cannot apply Hahn-Banach because you need that $\delta'$ is continuous (a bounded linear functional) with respect to the seminorm on $D$ inherited from its inclusion in $C$. That is, you need to use the $C^0$ seminorm, which is insufficient to control the value of $\delta'$. – Willie Wong Dec 23 '10 at 20:38
Thanks Willie. Somehow I thought $\sup_x f(x) \leq \sup_{x,\alpha} D^{\alpha}f(x)$ would take care of it. I'm not sure what I was (not) thinking... I've also seen the Hahn-Banach theorem stated a number of times without the continuous assumption, so this confused me. – David Ferrone Dec 24 '10 at 13:54

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